Given equation of ellipse is
Here,
$$
\begin{array}{r}
\frac{x^{2}}{144}+\frac{y^{2}}{25}=1 \\
a^{2}=144 \text { and } b^{2}=25
\end{array}
$$
Now.
$$
e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{25}{144}}
$$
$$
=\sqrt{\frac{119}{144}}=\frac{\sqrt{119}}{12}
$$
Foci of an ellipse $=(\pm a e, 0)$
$$
\begin{array}{l}
=\left(\pm 12 \times \frac{\sqrt{119}}{12}, 0\right) \\
=(\pm \sqrt{119}, 0)
\end{array}
$$
Since, the circle with centre $(0, \sqrt{2})$ and passing through foci $(\pm \sqrt{119}, 0)$ of the ellipse.
$\therefore$ Radius of circle $=\sqrt{(\sqrt{119}-0)^{2}+(0-\sqrt{2})^{2}}$
$$
=\sqrt{119+2}=\sqrt{121}=11
$$